blocks|key|1429624|text|如果矩形与另一个矩形重叠，则会发现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1429625|public+boolean+overlaps+(Rectangle+r)+{
++++return+x+<+r.x+%2B+r.width+&&+x+%2B+width+>+r.x+&&+y+<+r.y+%2B+r.height+&&+y+%2B+height+>+r.y;
}|code-block|syntax|javascript|1429626|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This will find if the rectangle is overlapping another rectangle:

<pre><code>public boolean overlaps (Rectangle r) {
 return x &lt; r.x + r.width &amp;&amp; x + width &gt; r.x &amp;&amp; y &lt; r.y + r.height &amp;&amp; y + height &gt; r.y;
}
</code></pre>

blocks|key|2147288|text|背景：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2147289|矩形只可由其对角线之一定义。|2147290|假设第一个矩形的对角线是(x1，y1)到(x2，y2)|2147291|另一个矩形的对角线是(x3，y3)到(x4，y4)|2147292|​|2147293|📷|atomic|offset|length|2147294|2147295|程序：|2147296|现在，如果这四个条件中的任何一个是正确的，我们可以得出结论，矩形不是重叠的：|2147297|x3+>+x2+(OR)|ordered-list-item|2147298|y3+>+y2+(OR)|2147299|x1+>+x4+(OR)|2147300|y1+>+y4|2147301|，否则，它们会重叠！|style|BOLD|2147302|2147303|2147304|2147305|另一种选择是：|2147306|矩形在下列情况下重叠|2147307|(x1+<+x4)+&&+(x3+<+x2)+&&+(y1+<+y4)+&&+(y3+<+y2)|code-block|syntax|javascript|2147308|Leetcode的示例解决方案：https://leetcode.com/problems/rectangle-overlap/discuss/468548/Java-check-if-two-rectangles-overlap-at-any-point|2147309|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/c40882d5c06ca76b93283095fb465637.jpg|imageAlt|1|https://developer.qcloudimg.com/http-save/yehe-900000/8e91490e749c38916bbd8c0da9dce570.jpg|2|LINK|MUTABLE|url|https://leetcode.com/problems/rectangle-overlap/discuss/468548/Java-check-if-two-rectangles-overlap-at-any-point^0|0|0|0|0|0|0|1|0|0|0|0|0|0|0|0|0|0|A|0|0|0|1|1|0|0|0|0|0|G|34|2|0^^$0|@$1|2|3|4|5|6|7|20|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|21|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|22|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|23|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|24|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|25|8|@]|9|@$M|26|N|27|1|28]]|A|$]]|$1|O|3|I|5|6|7|29|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|2A|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|2B|8|@]|9|@]|A|$]]|$1|T|3|U|5|V|7|2C|8|@]|9|@]|A|$]]|$1|W|3|X|5|V|7|2D|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|V|7|2E|8|@]|9|@]|A|$]]|$1|10|3|11|5|V|7|2F|8|@]|9|@]|A|$]]|$1|12|3|13|5|6|7|2G|8|@$M|2H|N|2I|14|15]]|9|@]|A|$]]|$1|16|3|I|5|6|7|2J|8|@]|9|@]|A|$]]|$1|17|3|K|5|L|7|2K|8|@]|9|@$M|2L|N|2M|1|2N]]|A|$]]|$1|18|3|I|5|6|7|2O|8|@]|9|@]|A|$]]|$1|19|3|1A|5|6|7|2P|8|@]|9|@]|A|$]]|$1|1B|3|1C|5|6|7|2Q|8|@]|9|@]|A|$]]|$1|1D|3|1E|5|1F|7|2R|8|@]|9|@]|A|$1G|1H]]|$1|1I|3|1J|5|6|7|2S|8|@]|9|@$M|2T|N|2U|1|2V]]|A|$]]|$1|1K|3|-4|5|6|7|2W|8|@]|9|@]|A|$]]]|1L|$1M|$5|1N|1O|1P|A|$1Q|1R|1S|-4]]|1T|$5|1N|1O|1P|A|$1Q|1U|1S|-4]]|1V|$5|1W|1O|1X|A|$1Y|1Z]]]]

<h3>Background:</h3>

A rectangle can be defined by just one of its diagonal. 
Let's say the first rectangle's diagonal is (x1, y1) to (x2, y2) 
And the other rectangle's diagonal is (x3, y3) to (x4, y4) 

<a href="https://i.stack.imgur.com/gLsbi.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/gLsbi.jpg" alt="Sample"></a>

<h3>Proceeding:</h3>

Now, if any of these 4 conditions is true, we can conclude that the rectangles are not overlapping:

<ol>
<li>x3 > x2 (OR)</li>
<li>y3 > y2 (OR)</li>
<li>x1 > x4 (OR)</li>
<li>y1 > y4</li>
</ol>

 
Otherwise, they overlap!

<a href="https://i.stack.imgur.com/bDtAF.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/bDtAF.jpg" alt="enter image description here"></a>

<h3>Alternatively:</h3>

The rectangles overlap if

<pre><code>(x1 &lt; x4) &amp;&amp; (x3 &lt; x2) &amp;&amp; (y1 &lt; y4) &amp;&amp; (y3 &lt; y2)
</code></pre>

 
Sample solution on Leetcode:
<a href="https://leetcode.com/problems/rectangle-overlap/discuss/468548/Java-check-if-two-rectangles-overlap-at-any-point" rel="noreferrer">https://leetcode.com/problems/rectangle-overlap/discuss/468548/Java-check-if-two-rectangles-overlap-at-any-point</a>

blocks|key|2425124|text|我会创建矩形对象，然后使用Rectangle.intersects和Rectangle.contains方法来确定它们是相交的，还是其中一个包含另一个。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2425125|因为你有一个很大的矩形，那就是选择矩形，这比我想象的更容易。运行Rectangle.contains，对于所有未包含的矩形，运行Rectangle.intersects，您就有了所需的内容。|2425126|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/8/docs/api/java/awt/Rectangle.html^0|D|K|Y|I|4|4|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@$9|U|A|V|1|W]]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

I would make <a href="http://docs.oracle.com/javase/8/docs/api/java/awt/Rectangle.html" rel="nofollow noreferrer">Rectangle objects</a> and then use the <code>Rectangle.intersects</code> and <code>Rectangle.contains</code> methods to determine if they intersect or if one contains the other.

Since you have one big rectangle, that is the selection rectangle, this is even easier than I thought. Run Rectangle.contains, and for all rectangles that aren't contained, run Rectangle.intersects, and you have what you are seeking.

blocks|key|2147322|text|以下是另一个更简单的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2147323|++++//+Left+x+
++++int+leftX+=+Math.max(x1,+x3);
++++//+Right+x
++++int+rightX+=+Math.min(x2,+x4);
++++//+Bottom+y
++++int+botY+=+Math.max(y1,+y3);
++++//+TopY
++++int+topY+=+Math.min(y2,+y4);

++++if+(rightX+>+leftX+&&+topY+>+botY)
+++++++return+true;|code-block|syntax|javascript|2147324|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's another simpler solution:

<pre><code> // Left x 
 int leftX = Math.max(x1, x3);
 // Right x
 int rightX = Math.min(x2, x4);
 // Bottom y
 int botY = Math.max(y1, y3);
 // TopY
 int topY = Math.min(y2, y4);

 if (rightX &gt; leftX &amp;&amp; topY &gt; botY)
 return true;
</code></pre>

blocks|key|2147131|text|我对gps坐标系中的多边形有一个通用的实现，对于矩形(简单的多边形)来说，这可能有点过分；但是它会工作的。如果出于任何原因您不想使用AWT，那么将这种方法调整到您的用法应该是相当简单的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2147132|重叠法(+https://github.com/jillesvangurp/geogeometry/blob/master/src/main/java/com/jillesvangurp/geo/GeoGeometry.java#L753+)|offset|length|2147133|我所做的只是简单地检查多边形是否有其他多边形包含的任何点。|2147134|对于点的多边形包含，我有一个简单的算法，它遍历多边形的边缘，以检查点是在O(n)内还是外部。对于矩形来说，运行起来应该很便宜。|2147135|这种方法的好处在于，它适用于任意矩形，也适用于旋转矩形或更复杂的形状。|2147136|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/jillesvangurp/geogeometry/blob/master/src/main/java/com/jillesvangurp/geo/GeoGeometry.java#L753^0|0|5|36|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@$D|V|E|W|1|X]]|A|$]]|$1|F|3|G|5|6|7|Y|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|Z|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|10|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S]]]]

I have a generic implemententation for polygons in the gps coordinate system, which may be a bit overkill for rectangles (which are simple polygons); but it will work. It should be fairly straightforward to adapt the approach to your usecase if for whatever reason you don't want to use AWT.

<a href="https://github.com/jillesvangurp/geogeometry/blob/master/src/main/java/com/jillesvangurp/geo/GeoGeometry.java#L753" rel="nofollow">https://github.com/jillesvangurp/geogeometry/blob/master/src/main/java/com/jillesvangurp/geo/GeoGeometry.java#L753</a> (overlap method)

What I do there is simply check if the polygons have any points that are contained by the other polygon.

For polygon containment of points, I have a simple algorithm that walks the edges of the polygon to check if the point is inside or outside O(n). For rectangles it should be cheap to run.

The nice thing about this approach it will work for any rectangles and also rotated rectangles or more complex shapes.

blocks|key|2425295|text|如果下列条件之一为真，两个矩形不重叠。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2425296|(+1)一个矩形位于另一个矩形的顶部边缘之上。|2425297|(+2)一个矩形位于另一个矩形的左边缘左侧。|2425298|注意，矩形可以由两个坐标表示，即左上角和右下角。因此，我们主要给出了四个坐标。|2425299|l1:第一个矩形的左上角坐标。|2425300|r1:第一个矩形的右下角坐标。|2425301|l2:第二个矩形的左上角坐标。|2425302|r2:第二个矩形的右下角坐标。|2425303|class+Point
{
++++int+x,+y;
};

//+Returns+true+if+two+rectangles+(l1,+r1)+and+(l2,+r2)+overlap
bool+doOverlap(Point+l1,+Point+r1,+Point+l2,+Point+r2)
{
++++//+If+one+rectangle+is+on+left+side+of+other
++++if+(l1.x+>+r2.x+%7C%7C+l2.x+>+r1.x)
++++++++return+false;

++++//+If+one+rectangle+is+above+other
++++if+(l1.y+<+r2.y+%7C%7C+l2.y+<+r1.y)
++++++++return+false;

++++return+true;
}|code-block|syntax|javascript|2425304|entityMap^0|F|3|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|14|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|15|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|16|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|17|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|18|8|@]|D|@]|E|$]]|$1|R|3|S|5|6|7|19|8|@]|D|@]|E|$]]|$1|T|3|U|5|V|7|1A|8|@]|D|@]|E|$W|X]]|$1|Y|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|Z|$]]

Two rectangles do not overlap if one of the following conditions is true. 
1) One rectangle is above top edge of other rectangle. 
2) One rectangle is on left side of left edge of other rectangle.

Note that a rectangle can be represented by two coordinates, top left and bottom right. So mainly we are given following four coordinates. 
l1: Top Left coordinate of first rectangle. 
r1: Bottom Right coordinate of first rectangle. 
l2: Top Left coordinate of second rectangle. 
r2: Bottom Right coordinate of second rectangle. 

<pre><code>class Point
{
 int x, y;
};

// Returns true if two rectangles (l1, r1) and (l2, r2) overlap
bool doOverlap(Point l1, Point r1, Point l2, Point r2)
{
 // If one rectangle is on left side of other
 if (l1.x &gt; r2.x || l2.x &gt; r1.x)
 return false;

 // If one rectangle is above other
 if (l1.y &lt; r2.y || l2.y &lt; r1.y)
 return false;

 return true;
}
</code></pre>

blocks|key|2425320|text|该类假定排序为left<=right、top<=bottom、x1<=x2、y1<=y2|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2425321|public+class+Rect
{
int+left,+right,+bottom,+top;

Rect(int+left,+int+top,+int+right,+int+bottom)
{
++++this.left+=+left;
++++this.right+=+right;
++++this.top+=+top;
++++this.bottom+=+bottom;
}

boolean+overlap(int+x1,+int+y1,+int+x2,+int+y2)
{
++++//+if+one+rectangle+is+to+the+left+or+right,+then+there+can+be+no+overlap
++++if(x2+<+left+%7C%7C+right+<+x1)
++++++++return+false;

++++//+the+x+values+overlap,+but+the+y+values+may+still+lie+outside+the+rectangle

++++//+if+one+rectangle+is+above+or+below,+then+there+can+be+no+overlap
++++if(y2+<+top+%7C%7C+bottom+<+y1)
++++++++return+false;

++++//+otherwise+we+must+overlap+!
++++return+true;++++++++
}
}|code-block|syntax|javascript|2425322|entityMap^0|7|B|J|B|V|6|12|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

This class assumes the ordering <code>left&lt;=right</code>, <code>top&lt;=bottom</code>, <code>x1&lt;=x2</code>, <code>y1&lt;=y2</code>:

<pre><code>public class Rect
{
int left, right, bottom, top;

Rect(int left, int top, int right, int bottom)
{
 this.left = left;
 this.right = right;
 this.top = top;
 this.bottom = bottom;
}

boolean overlap(int x1, int y1, int x2, int y2)
{
 // if one rectangle is to the left or right, then there can be no overlap
 if(x2 &lt; left || right &lt; x1)
 return false;

 // the x values overlap, but the y values may still lie outside the rectangle

 // if one rectangle is above or below, then there can be no overlap
 if(y2 &lt; top || bottom &lt; y1)
 return false;

 // otherwise we must overlap !
 return true; 
}
}
</code></pre>

blocks|key|1429827|text|java.awt.Rectangle有一个内置的intersects方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1429828|import+java.awt.Rectangle;
//+...
Rectangle+r1+=+new+Rectangle(
++++0+/*+top+left+x+*/,+0+/*+top+left+y+*/,+
++++5+/*+width+*/,+7+/*+height+*/
);
Rectangle+r2+=+new+Rectangle(4,+5,+3,+3);
System.out.println(r1.intersects(r2));+//+true|code-block|syntax|javascript|1429829|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/awt/Rectangle.html|1|https://docs.oracle.com/javase/8/docs/api/java/awt/Rectangle.html#intersects-java.awt.Rectangle-^0|0|I|O|A|0|I|0|O|A|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]|$9|12|A|13|1|14]]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]]]

<a href="https://docs.oracle.com/javase/8/docs/api/java/awt/Rectangle.html" rel="nofollow noreferrer"><code>java.awt.Rectangle</code></a> has a built-in <a href="https://docs.oracle.com/javase/8/docs/api/java/awt/Rectangle.html#intersects-java.awt.Rectangle-" rel="nofollow noreferrer"><code>intersects</code></a> method for this.
<pre><code>import java.awt.Rectangle;
// ...
Rectangle r1 = new Rectangle(
 0 /* top left x */, 0 /* top left y */, 
 5 /* width */, 7 /* height */
);
Rectangle r2 = new Rectangle(4, 5, 3, 3);
System.out.println(r1.intersects(r2)); // true
</code></pre>

I have multiple rectangles and one special rectangle: the selection rect.
I want to check for each rectangle if the rectangle contains at least one point which is inside the selection rectangle.
Here is an image for clarity:

<img src="https://i.stack.imgur.com/OS2GT.png" alt="Selection example">

Java check if two rectangles overlap at any point

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有多个矩形和一个特殊的矩形:选择矩形。我想要检查每个矩形，如果矩形包含至少一个点在选择矩形内。这是一幅清晰的图像：​​

问Java检查两个矩形是否在任意一点重叠
EN

回答 8

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问Java检查两个矩形是否在任意一点重叠
EN